\(\left\{{}\begin{matrix}x^2+y^2=2\left(1\right)\\3y^2+4xy+x+2y=10\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right)\Leftrightarrow x^2+4xy+4y^2+x+2y=12\)
\(\Leftrightarrow\left(x+2y\right)^2+\left(x+2y\right)-12=0\)
\(\Leftrightarrow\left(x+2y-3\right)\left(x+2y+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2y-3=0\\x+2y+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3-2y\\x=-2y-4\end{matrix}\right.\)
Với \(x=3-2y\) :
\(\left(1\right)\Leftrightarrow y^2+\left(3-2y\right)^2=2\)
\(\Leftrightarrow5y^2-12y+7=0\)
\(\Leftrightarrow\left(y-1\right)\left(5y-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=\frac{7}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{5}\end{matrix}\right.\)
Với \(x=-2y-4\) :
\(\left(1\right)\Leftrightarrow y^2+\left(-2y-4\right)^2=2\)
\(\Leftrightarrow5y^2+16y+14=0\)
\(\Delta'=8-60=-62< 0\)
\(\Rightarrow PTVN\)
Vậy \(\left[{}\begin{matrix}\left(x;y\right)=\left(1;1\right)\\\left(x;y\right)=\left(\frac{1}{5};\frac{7}{5}\right)\end{matrix}\right.\)